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Given that $\sum a_{n}$ converges $\left(a_{n}>0\right) ;$ Then $(\sum a_{n}^{3} \sin n)$ is


My approach:

Since, $\sum a_{n}$ converges, we have $\lim _{n \rightarrow \infty} n \cdot a_{n}$ converges.

i.e. $\left|n \cdot a_{n}\right| \leq 1$ for $n \geq K(\text { say })$

$\Rightarrow n \cdot a_{n}<1 \quad\left[\because a_{n}>0\right]$

$\Rightarrow a_{n}<\frac{1}{n}$

$\therefore a_{n}^{3}<\frac{1}{n^{3}}$

$\Rightarrow a_{n}^{3} \sin n \leq \frac{1}{n^{3}} \sin n \leq \frac{1}{n^{3}}$

$\Rightarrow \sum a_{n}^{3} \sin n \leq \sum \frac{1}{n^{3}}$

$\because \mathrm{RHS}$ converges so LHS will also converge.

Any other better approach will be highly appreciated and correct me If I am wrong

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    $\begingroup$ It is not true that if $\sum_{n=1}^{\infty} a_n$ converges then $a_n\leq 1/n$ for all sufficiently large $n$. You can have $a_n>1/n$ infinitely often. $\endgroup$ – Michael Jul 13 '20 at 19:34
  • $\begingroup$ so my approach seems wrong $\endgroup$ – user791682 Jul 13 '20 at 19:34
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    $\begingroup$ Your assertion Since, $\sum a_{n}$ converges, we have $\lim _{n \rightarrow \infty} n \cdot a_{n}$ converges. is wrong in general. It is true if $\{a_n\}$ is supposed to be positive decreasing. $\endgroup$ – mathcounterexamples.net Jul 13 '20 at 19:35
  • $\begingroup$ You want to compare $\sum_{n=m}^{\infty} a_n^3 \sin(n)$ to $\sum_{n=m}^{\infty} a_n$. $\endgroup$ – Michael Jul 13 '20 at 19:36
  • $\begingroup$ Is $a_n\ge 0$ an assumption? If not, take $a_n=\frac{\sin(n)}{n^{1/3}}$. Then, we have $\sum_n a_n$ converges. But does $\sum_n a_n^3\sin(n)=\sum_n \frac{\sin^4(n)}{n}$ converge? $\endgroup$ – Mark Viola Jul 13 '20 at 20:35
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As $\{a_n\}$ is a positive sequence such that $\sum a_n$ converges, we have $0 \le a_n \le 1$ for $n$ large enough, say $n \ge M$.

Then for $n \ge M$

$$0 \le \vert a_n^3 \sin n \vert \le a_n^3 \le a_n$$

Hence $\sum a_n^3 \sin n$ converges absolutely.

Also a series $\sum a_n$ can be convergent while the sequence $\{n a_n\}$ diverges. Consider for example $a_n$ equals to $0$ if $n$ is not a square and equal to $1/n$ otherwise.

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For $a_n\geq 0$ the sum converges as shown by @mathcounterexamaples.net

For general $a_n$ the statement is not true. The (deleted) counter-example given by @Mark Viola indeed is a counter-example. Taking $a_n=\sin(n)/n^{1/3}$ it suffices to show that $\sum_{n\geq 1} a_n^3 \sin(n) =\sum_{n\geq 1} \frac{\sin^4 n}{n} =+\infty$. To see this note that $n$ is asymptotically equidistributed mod $2\pi$ which implies that for any continuous function $f\in C([-1,1])$, $ \lim_{N\rightarrow +\infty} \frac{1}{N} \sum_{n=1}^N f(\sin(n)) = \frac{1}{2\pi} \int_0^{2\pi} f(\sin(t))\; dt$. In particular, as $N\rightarrow +\infty$: $$ M_N = \frac{1}{N} \sum_{n=1}^N \sin^4(n) \rightarrow \frac{1}{2\pi} \int_0^{2\pi} \sin^4(t)\; dt = \frac{3}{8}.$$ Proceeding as in Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$ find a strictly increasing sequence $N_k$, $k\geq 1$ so that each $M_{N_k}\geq \frac{1}{4}$ and $8 N_k \leq N_{k+1}$. Then $$ \sum_{n=1+N_k}^{N_{k+1}} \frac{\sin^4(n)}{n} \geq \frac{1}{N_{k+1}} \sum_{n=1+N_k}^{N_{k+1}} \sin^4(n) \geq M_{N_{k+1}} -\frac{N_k}{N_{k+1}} \geq \frac{1}{4} - \frac{1}{8}= \frac{1}{8} $$ Summing over $k$ implies the divergence of $\sum_{n\geq 1} \frac{\sin^4(n)}{n}$.

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  • $\begingroup$ Regarding showing $\sum_n \sin^4(n)/n$: an easier way is to write $8\sin^4(n)=3-4\cos(2x)+\cos(4x)$. Then $\sum_n \cos(a n)/n$ converges iff $a$ isn't a multiple of $2\pi$ and the harmonic series diverges. $\endgroup$ – FearfulSymmetry Jul 24 '20 at 19:21
  • $\begingroup$ @Integrand Good point. $\endgroup$ – H. H. Rugh Jul 24 '20 at 20:04

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